# proof of Cauchy-Schwarz inequality

If $a$ and $b$ are linearly dependent, we write $\boldsymbol{b}=\lambda\boldsymbol{a}$. So we get:

 $\langle\boldsymbol{a},\lambda\boldsymbol{a}\rangle^{2}=\lambda^{2}\langle% \boldsymbol{a},\boldsymbol{a}\rangle^{2}=\lambda^{2}||\boldsymbol{a}||^{4}=||% \boldsymbol{a}||^{2}||\boldsymbol{b}||^{2}.$

So we have equality if $\boldsymbol{a}$ and $\boldsymbol{b}$ are linearly dependent. In the other case we look at the quadratic function

 $||x\cdot\boldsymbol{a}+\boldsymbol{b}||^{2}=x^{2}||\boldsymbol{a}||^{2}+2x% \langle\boldsymbol{a},\boldsymbol{b}\rangle+||\boldsymbol{b}||^{2}.$

This function is positive for every real $x$, if $\boldsymbol{a}$ and $\boldsymbol{b}$ are linearly independent. Thus it has no real zeroes, which means that

 $\langle\boldsymbol{a},\boldsymbol{b}\rangle^{2}-||\boldsymbol{a}||^{2}||% \boldsymbol{b}||^{2}$

is always negative. So we have:

 $\langle\boldsymbol{a},\boldsymbol{b}\rangle^{2}<||\boldsymbol{a}||^{2}||% \boldsymbol{b}||^{2},$

which is the Cauchy-Schwarz inequality if $\boldsymbol{a}$ and $\boldsymbol{b}$ are linearly independent.

Title proof of Cauchy-Schwarz inequality ProofOfCauchySchwarzInequality 2013-03-22 12:34:42 2013-03-22 12:34:42 mathwizard (128) mathwizard (128) 6 mathwizard (128) Proof msc 15A63